Is it possible to work out the centre of an ellipse?

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Homework Help Overview

The discussion revolves around determining the eccentric angle of an ellipse defined by the equation x² + 9y² = 13 at a specific point (2,1). Participants are exploring the implications of the ellipse's center and the relationship between the point and the axes of the ellipse.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand how to calculate the eccentric angle, initially considering the angle as arctan(1/2) based on a standard Cartesian interpretation. Other participants clarify that the ellipse is centered at the origin and emphasize the need to account for the ellipse's axes in the calculation of the eccentric angle.

Discussion Status

Participants are actively engaging with the problem, with some providing clarifications regarding the definition of the eccentric angle and its dependence on the ellipse's geometry. There is an ongoing exploration of the correct approach to the problem, but no consensus has been reached.

Contextual Notes

There is some confusion regarding the interpretation of the eccentric angle and the relationship between the point (2,1) and the ellipse's axes. Participants are also referencing external resources for further clarification.

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Is it possible to work out the centre of an ellipse?
The question asks for the eccentric angle of the ellipse with the equation x²+9y²=13 at point (2,1)...
I have no idea how to get this, I know that the angle would be arctan(1/2) if the ellipse was centred at (0,0)

Thanks
 
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But the ellipse IS centered at the origin. It asks for the eccentric angle between the x-axis and the line joining (0,0) and (2,1).

edit: PS: The eccentric angle is not simply arctan(y/x), you have to take the axes of the ellipse into account too!

- Kamataat
 
Last edited:
Kamataat said:
But the ellipse IS centered at the origin. It asks for the eccentric angle between the x-axis and the line joining (0,0) and (2,1).

edit: PS: The eccentric angle is not simply arctan(y/x), you have to take the axes of the ellipse into account too!

- Kamataat
I am somewhat confused. Surely the line joining (0,0) and (2,1) makes arctan (.5) of an angle with the x axis.:confused:
 
Thanks a lot :biggrin:
 

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